The Principles of Switching Power Conversion--Evolution of Switching Topologies

Controlling the Induced Voltage Spike by Diversion through a Diode

We realize that our "problem" with using an inductor is two-fold: either we are going to end up with near-infinite induced voltage spikes, as shown in Fgr. 6 and Fgr. 7, or if we do somehow manage to control the induced voltage to some finite level, the equation V = L dI/dt tells us we could very well end up with near-infinite currents (staircasing).

And further, coming to think of it, our basic purpose is still not close to being fulfilled - we still don't know how to derive any useful power from our circuit! Luckily, all the above problems can be solved in one stroke! And in doing so, we will arrive at our very first 'switching topology.' Let's now see how that comes about.

We recollect from Fgr. 6 that the spike of induced voltage at switch turn-off occurs only because the current (previously flowing in the inductor) was still demanding a path along which to flow - and somehow unknowingly, we had failed to provide any. Therefore nature, in search of the "weakest link," found this in the switch itself - and produced an arc across it, to try and move the current across anyway.

But suppose we consciously provide a "diversionary path"? Then there would be no problem turning the switch OFF and stopping the inductor current flowing through the switch - because it could continue to flow via this alternate route. The inductor would then no longer "complain" in the form of a dangerous voltage spike. Thereafter, perhaps we can even re-route the current back into the switch when it turns ON again. Finally, we can perhaps even repeat the ON-OFF-ON-OFF process indefinitely, at a certain switching frequency.

In Fgr. 10 we have created such an alternate path. We will see that the way the diode is pointed, this path can come into play automatically, and only when the switch turns OFF.

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Fgr. 10: Providing a "Diversion" for the Inductor Current through a Diode

Example:

VIN = 12V is the applied dc voltage VD = 0.5V is the forward drop across diode

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Just to make things clearer, we have used some sample numbers in Fgr. 10. We have taken the applied input voltage to be 12 V and assumed a typical Schottky diode forward drop of 0.5 V. Note that we are assuming a "perfect" switch here (no forward drop), for the sake of simplicity. We make the following observations:

+++ When the switch is ON (closed), the voltage at the upper end of the inductor L is at 12 V and the lower end is at 0 V ('ground'). So the diode is reverse-biased and does not conduct. Energy is then being built up in the inductor by the applied dc voltage source.

The magnitude of the voltage applied across the inductor during the on-time of the switch (i.e. 'VON') is equal to 12 V.

+++ When the switch turns OFF (open), an alternate path is available for the inductor current to flow - through the diode. And we can be sure that "nature" (in our case the "induced voltage") will attempt to exploit this path - by forcing the diode to conduct. But for that, the diode must get 'forward-biased,' that is, its anode must get to a voltage 0.5 V higher than the cathode. But the anode is being held at ground (0 V rail). Therefore, the cathode must fall to -0.5 V.

The magnitude of the voltage applied across the inductor during the off-time of the switch (i.e. 'VOFF') is equal to 0.5 V.

+++ Note that the induced voltage during the switch off-time has had its polarity reversed.

+++ The rate of rise of the current (in the inductor and switch) during the on-time is equal to VON/L. And during the off-time, the current ramps down (much more slowly), at a rate of VOFF/L (in the inductor and diode).

+++ Yes, if we wait long enough the inductor current will finally ramp down to zero (inductor 'reset'). But if we don't wait, and turn the switch back ON again, the current will again start to ramp up (staircasing), as shown in Fgr. 10.

+++ Note that both the switch and diode currents have a "choppy" waveform - since one takes over where the other left off. This is in fact always true for any switching power converter (or topology).

Summarizing: We see that having provided a diversionary path for the current, the inductor isn't 'complaining' anymore, and there is no uncontrolled induced voltage spike anymore.

But we certainly have now ended up with a possible problem of escalating currents. And come to think of it, neither do we have a useful output rail yet, which is what we are basically looking to do finally. In fact, all that we are accomplishing in Fgr. 10 is dissipating some of the stored energy built-up in the inductor during the on-time, within the diode during the off-time.

Achieving a Steady State and Deriving Useful Energy

We realize, that to prevent staircasing, we need to somehow induce volt-seconds balance. Yes, as mentioned, we could perhaps wait long enough before turning the switch ON again. But that still won't give us a useful output rail.

To finally solve all our problems in one go, let us take a hint from our "natural world of voltages." Since we realize we are looking for an output dc voltage rail, isn't it natural to use a capacitor somewhere in the circuit of Fgr. 10? Let us therefore now interpose a capacitor in series with the diode, as shown in Fgr. 11. If we do that, the diode (freewheeling) current would charge the capacitor up - and hopefully the capacitor voltage would eventually reach a steady level 'VO'! Further, since that would increase the voltage drop appearing across the inductor during the off-time (VOFF), it would increase the rate at which the inductor current can ramp down - which we recognize was the basic problem with the circuit in Fgr. 10. So we are finally seeing light at the end of the tunnel - by making VOFF comparable to VON, we are hoping to achieve volt-seconds balance expressed by VON × tON = VOFF × tOFF.

In Fgr. 11 the current escalates initially, but then after several cycles, automatically levels out, in what is clearly a steady state. That is because every cycle the capacitor charges up, it progressively increases the slope of the down-ramp, eventually allowing the converter to settle down naturally into the basic condition fliON = fliOFF = fli. And once that is achieved, it’s self-sustaining! We also have a useful rail now - available across the output capacitor, from which we can draw some stored energy. So we have shown a dc current passing through to the load by the dashed arrows in Fgr. 11.

In fact, this is our very first switching topology - the buck-boost topology.

Note: Under the abnormal condition of an output short For example, Fgr. 11 effectively reduces to Fgr. 10! Therefore, to protect the converter under such conditions, a current limit is required.

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Example:

VIN

= 2V is the applied dc voltage VD

= 0.5V is the forward drop across diode VO

= 5V is the final value of voltage across output capacitor

IL Inductor Current (startup simplified)

Fgr. 11: Evolution of the Buck-boost Topology

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The Buck-boost Converter

To understand Fgr. 11 better, we are actually going to work backward from here. So let us assume we have achieved a steady state - and therefore the output capacitor too has reached a steady value of say, 5 V. Let us now find the conditions needed to make that a reality.

In Fgr. 11 the slope of the rising ramp is unchanged every cycle, being equal to VIN/L.

The slope of the falling ramp is initially VD/L, where "VD" is the drop across the diode. So from the inductor equation, initially fliON > fliOFF. Thus the current starts to staircase. But the magnitude of the slope of the falling ramp, and therefore fliOFF, keeps getting larger and larger as the capacitor charges up. Eventually we will reach a steady state defined by

fliOFF = fliON. At that moment, the volt-seconds law applies.

....

Using the numbers of the example, we get

....

We see that a 5 V output is possible only if we have been switching with a constant ratio between the switch ON and switch OFF time, as given by....

So to get the volt-seconds to balance out for this case (5 V output and a 12 V input), we have to make the off-time 2.18 times larger than the on-time. Why so? Simply because the voltage during the on-time (across the inductor) is larger by exactly the same proportion: 12 V during the on-time as compared to 5.5 V during the off-time. Check: 12/5.5 = 2.18.

The duty cycle (assuming CCM) is therefore equal to ...

Now, had we taken a semiconductor switch instead of a mechanical one, we would have had a non-zero forward voltage drop, of say "VSW." This forward drop effectively just subtracts from the applied dc input during the on-time. So, had we done the above calculations symbolically, we would get:

VON = VIN - VSW (Buck-boost)

… and …

VOFF = VO + VD (Buck-boost)

Then, from the volt-seconds law ... We thus get the duty cycle...If the switch and diode drops are small as compared to the input and output rails, we can simply write ...

We can also write the relationship between the input and output as follows ...

Note that some other easily derivable, and convenient relationships to remember are ...

..... where we have defined D' = 1 - D as the 'duty cycle of the diode,' since the diode is conducting for the remainder of the switching cycle duration (in CCM).

Ground-referencing

Our Circuits We need to clearly establish what is referred to as the 'ground' rail in any dc-dc switching topology. We know that there are two rails by which we apply the dc input voltage (current goes in from one and returns from the other). Similarly, there are also two output rails. But all practical topologies generally have one rail that is common to both the input and the output. It’s this common rail, that by convention, is called the system 'ground' in dc-dc converter applications.

However, there is yet another convention in place - the ground is also considered to be "0 V" (zero volts).

The Buck-boost Configurations

In Fgr. 12 the common (ground) rails have been highlighted in bold gray background.

We now realize that the buck-boost we presented in Fgr. 11 is actually a 'positive (input) to negative (output)' buck-boost. There is another possibility, as shown in the lower half of Fgr. 12. We have re-labeled its ground in accordance with the normal convention. Therefore this is a 'negative to positive buck-boost.' For either configuration, we see that whatever polarity is present at the input, it gets reversed at the output. Therefore, the buck-boost is often simply called an 'inverting' topology (though we should keep in mind that that allows for two different configurations).

Fgr. 12: The Two Configurations of the Buck-boost (inverting) Topology: Positive to Negative

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Fgr. 13: Analyzing the Buck-boost; inductor Note:

IL is the average Inductor Current IO is the average Load Current

BUCK-BOOST

Average Input Current equals Average Switch Current

Average Output Current is equal to average Diode Current

Minimum Switch Rating Minimum Diode Rating Minimum Inductor Rating

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The Switching Node

Very simply put - the "point of detour" for the inductor current, that is, between the switch and the diode, is called the 'switching node.' Current coming into this node from the inductor, can go either into the diode or the switch, depending upon the state of the switch.

Every dc-dc switching topology has this node (without it we would get the huge voltage spike we talked about!).

Since the current at this node needs to alternate between the diode and the switch, it needs to alternately force the diode to change state too (i.e. reverse-biased when the switch turns ON and forward-biased when the switch is OFF). So, the voltage at this node must necessarily be 'swinging.' An oscilloscope probe connected here (with its ground clip connected to the power supply ground, i.e. 0V), will always see a voltage waveform with "square edges." This is in fact very similar to the voltage across the inductor, except that it’s dc level-shifted by a certain amount, depending on the topology.

On a practical level, while designing the PCB (printed circuit board), we have to be cautious in not putting too much copper at the switching node. Otherwise it becomes an effective electric-field antenna, spewing radiated radio frequency interference all around.

The output cables can thereafter pick up the radiated noise and transmit it directly to the load.

Analyzing the Buck-boost

In Fgr. 13 we have drawn a line 'IL' through the geometric center of the ramp portion of the steady-state inductor current waveform. This is defined as the average inductor current.

The switch current also has an average value of IL during the interval tON. Similarly the average of the diode current is also IL, during tOFF. However, the switch and diode currents when averaged over the entire cycle (i.e. over both the ON and OFF durations) are by simple mathematics their respective weighted averages.

(Buck-boost)

...where D' is the duty cycle of the diode, that is, 1 - D. It’s also easy to visualize that for this particular topology, the average input current is equal to the average switch current. Further, as we will see below, the average diode current is equal to the load current. This is what makes the buck-boost topology quite different from the buck topology.

Properties of the Buck-boost

We now make some observations based on Fgr. 11, Fgr. 12, and Fgr. 13

+++ For example, a "positive to negative" buck-boost can convert 12 V to -5 V (step down) or 12 V to -15 V (step up). A "negative to positive" buck-boost can convert say -12 V to 5 V or 5 V to 15 V, and so on. The magnitude of the output voltage can thus be either smaller or larger (or equal to) the magnitude of the input voltage.

+++ When the switch is ON, energy is delivered only into the inductor by the input dc source (via the switch), and none of it passes through to the output.

+++ When the switch is OFF, only the stored energy of the inductor is pushed into the output (through the diode), and none comes directly from the input dc source.

+++ The above two observations make the buck-boost topology the only "pure flyback" topology around, in the sense that all the energy transferred from the input to the output, must have been previously stored in the inductor. No other topology shares this unique property.

+++ The current coming from the input capacitor (dc source) is "choppy," that is, pulsating. That is because, this current, combined with the steady dc current (IIN) coming in from the dc source, basically forms the switch current waveform (which we know is always choppy for any topology) (see Fgr. 9).

+++ Similarly, the current into the output capacitor is also choppy, because combined with the steady dc current into the load (I_OUT), it forms the diode current (which we know is always choppy for any topology) (see Fgr. 9).

+++ We know that heat dissipation is proportional to the square of the RMS current. And since choppy waveforms have high RMS values, the efficiency of a buck-boost is not very good. Also, there is generally a relatively high level of noise and ripple across the board. Therefore, the buck-boost may also demand much better filtering at its input, and often at its output too.

+++ Though current enters the output capacitor to charge it up when the switch turns ON, and leaves it to go into the load when the switch is OFF, the average capacitor current is always zero. In fact, any capacitor in 'steady state' must, by definition, have zero average current passing through it - otherwise it would keep charging or discharging until it too reaches a steady state, just like the inductor current.

Since the average current from the output capacitor is zero, therefore, for the buck-boost, the average diode current must be equal to the load current (where else can the current come from?). Therefore ...

1 - D (Buck-boost)

This is the relationship between the average inductor current and the load current. Note that in Fgr. 13, in the embedded table, we have asked for an inductor rated for 1.2 × 10/(1 - D). The factor "1.2" comes from the fact, that by typical design criteria, the peak of the inductor current waveform is about 20% higher than its average. So we need to look for an inductor rated at least for a current of 1.2 × IL.

Fgr. 14: Improperly Referenced Buck-boost Configurations: Buck-boost configurations with no proper Ground Reference

Why Three Basic Topologies Only?

There are certainly several ways to set up circuits using an inductor, which provide a "freewheeling path" too, for the inductor current. But some of these are usually disqualified simply because the input and output don’t share a common rail, and thus there is no proper ground reference available for the converter and the rest of the system. Two examples of such "working-but-unacceptable" converters are the buck-boost configurations shown in Fgr. 14. Compare these with Fgr. 12 to see what the problem is! However, note that if these were "front-end converters," the system ground could be established starting at the output of this converter itself, and may thus be acceptable.

Of the remaining ways, several are just "configurations" of a basic topology (like the two configurations in Fgr. 12). Among the basic topologies, we actually have just three - the buck, the boost, and the buck-boost. Why only three? That is because of the way the inductor is connected. Note that with proper ground-referencing in place, there are only three distinct rails possible - the input, the output, and the (common) ground. So if one end of the inductor is connected to the ground, it becomes a buck-boost! On the other hand, if it’s connected to the input, it becomes a boost. And if connected to the output, it becomes a buck. See Fgr. 15.

The Boost Topology

Fgr. 15: Three Basic Topologies: Possible Only; IN

Fgr. 16: The (positive) Boost and Buck Topologies: Positive-to-Positive Boost

In Fgr. 16 we have presented the schematic of the boost topology. The direct and the freewheeling paths are indicated therein. In Fgr. 17, we have the corresponding analysis, including the key waveforms.

We now make some observations:

+++ For example, a "positive to positive" boost can convert 12 V to 50 V. A "negative to negative" boost would be able to convert say -12Vto -50 V. The magnitude of the output voltage must therefore always be larger than the magnitude of the input voltage. So a boost converter only steps-up, and also does not change the polarity.

+++ When the switch is ON, energy is delivered only into the inductor by the input dc source (via the switch), and none of it passes through to the output.

+++ When the switch is OFF, the stored energy of the inductor is pushed into the output (through the diode). But some of it also comes from the input dc source.

+++ The current coming from the input capacitor (dc source) is "smooth," since it’s in series with the inductor (which prevents sudden jumps in current).

+++ However, the current into the output capacitor is "choppy," because combined with the steady dc current into the load (IOUT), it forms the diode current (which we know is always choppy for any topology) (see Fgr. 9).

+++ Since the average current from the output capacitor is zero, therefore, for the boost, the average diode current must be equal to the load current (where else can the current come from?). Therefore ...

This is the relationship between the average inductor current and the load current. Note that in Fgr. 17, in the embedded table, we have asked for an inductor rated for 1.2 × IO/(1 - D). The factor "1.2" comes from the fact, that by typical design criteria, the peak of the inductor current waveform is about 20% higher than its average. So we need to look for an inductor rated at least for a current of 1.2 × IL.

Let us analyze the boost topology in terms of the volt-seconds in steady state. We have ...

Fgr. 17: Analyzing the Boost

So, from the volt-seconds law ...

Performing some algebra on this to eliminate tOFF....

Finally, the 'duty cycle' of the converter D, which is defined as ...

D = tON/T (any topology)

...is the reciprocal of the preceding equation. So ....

We have thus derived the classical dc transfer function of a boost converter.

If the switch and diode drops are small as compared to the input and output rails, we can just write....

We can also write the relationship between the input and output as follows ...

Fgr. 18: Analyzing the Buck: Voltage at Switching Node

The Buck Topology

In Fgr. 16 we had also presented the schematic of the buck topology. The direct and the freewheeling paths are indicated therein. In Fgr. 18, we have the corresponding analysis, including the key waveforms.

We now make some observations:

+++ For example, a "positive to positive" buck can convert 12 V to 5 V. A "negative to negative" buck would be able to convert say -12Vto -5V.The magnitude of the output voltage must therefore always be smaller than the magnitude of the input voltage. So a buck converter only steps-down, and also does not change the polarity.

+++ When the switch is ON, energy is delivered to the inductor by the input dc source (via the switch). But some of it also passes through to the output.

+++ When the switch is OFF, the stored energy of the inductor is pushed into the output (through the diode). And none of it now comes from the input dc source.

+++ The current coming from the input capacitor (dc source) is "choppy." That is because, this current, combined with the steady dc current (IIN) coming in from the dc source, basically forms the switch current waveform (which we know is always choppy for any topology).

+++ However, the current into the output capacitor is "smooth," because it’s in series with the inductor (which prevents sudden jumps in current).

+++ Since the average current from the output capacitor is zero, therefore, for the buck, the average inductor current must be equal to the load current (where else can the current come from?). Therefore,

IL = IO (Buck)

This is the relationship between the average inductor current and the load current. Note that in Fgr. 18, in the embedded table, we have asked for an inductor rated for 1.2 × 10.

The factor "1.2" comes from the fact, that by typical design criteria, the peak of the inductor current waveform is about 20% higher than its average. So we need to look for an inductor rated at least for a current of 1.2 × IL.

Let us analyze the buck topology in terms of the volt-seconds in steady state. We have

VON = VIN - VSW - VO (Buck)

…and…

VOFF = VO - (-VD) = VO + VD (Buck)

As before, using the volt-seconds law and simplifying, we get the 'duty cycle' of the converter ...

We have thus derived the classical dc transfer function of a buck converter. If the switch and diode drops are small as compared to the input and output rails, we can just write ...

We can also write the relationship between the input and output as follows...

Advanced Converter Design

This should serve as an introduction to understanding and designing switching power converters. More details and worked examples can be found in the next section (titled "DC-DC Converter Design and Magnetics"). The reader can also at this point briefly scan Section 4 for some finer nuances of design. A full design table is also available in Sub-Section 2 for future reference.